Phase and rotation gates

A relative phase gate

$$P(\phi)=\begin{pmatrix}1&0\\0&e^{i\phi}\end{pmatrix}.$$

It leaves the magnitude of each amplitude unchanged. Therefore it does not change Z-basis probabilities immediately. S is $P(\pi/2)$, T is $P(\pi/4)$ and Z is $P(\pi)$.

The interference experiment places P between two H gates so you can observe its effect.

A rotation is related, but check its convention

$$R_z(\phi)=\begin{pmatrix}e^{-i\phi/2}&0\\0&e^{i\phi/2}\end{pmatrix}.$$

For a standalone one-qubit operation, $P(\phi)=e^{i\phi/2}R_z(\phi)$, so they differ only by global phase. Controlled versions require more care: a phase on one controlled branch can become a relative phase in the larger system.

The Y rotation is

$$R_y(\theta)=\begin{pmatrix}\cos(\theta/2)&-\sin(\theta/2)\\\sin(\theta/2)&\cos(\theta/2)\end{pmatrix}.$$

Applying it to zero gives amplitudes $\cos(\theta/2)$ and $\sin(\theta/2)$. At $\theta=2\pi/3$, the probabilities are $1/4$ and $3/4$.

Practise

Set the qubit explorer’s polar angle to 120° and phase to 0°. Compare it with the Y-rotation example. Then apply S and T and explain why the displayed Z probabilities stay fixed while the arrow moves.

Reference: IBM Quantum — Single systems.